How do you integrate #(x^2e^(x/2))dx#?

Question

4764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T20:11:30

#intx^2e^(x/2)dx=2e^(x/2)(x^2-4x+8)+C#

#I=intx^2e^(x/2)dx#

Use the substitution #t=x/2#. This implies that:

#{(2t=x),(2dt=dx),(4t^2=x^2):}#

Then:

#I=intx^2e^(x/2)dx=int(4t^2)(e^t)(2dt)=int8t^2e^tdt#

Now use integration by parts. This takes the form #intudv=uv-intvdu#. For the current integral, let:

#{(u=8t^2" "=>" "du=16tdt),(dv=e^tdt" "=>" "v=e^t):}#

So:

#I=uv-intvdu=8t^2e^t-int16te^tdt#

Again perform integration by parts with a knew #u# and #dv#:

#{(u=16t" "=>" "du=16dt),(dv=e^tdt" "=>" "v=e^t):}#

Hence:

#I=8t^2e^t-(uv-intvdu)#

#I=8t^2e^t-16te^t+int16e^tdt#

Note that #int16e^tdt=16inte^tdt=16e^t#:

#I=8t^2e^t-16te^t+16e^t+C#

#I=8e^t(t^2-2t+2)+C#

Returning to #x# from the substitution #t=x/2#:

#I=8e^(x/2)(x^2/4-x+2)+C#

#I=2e^(x/2)(x^2-4x+8)+C#